Gravitation

Class 11 NCERT Physics

NCERT

1   Answer the following:$\\$ (a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?$\\$ (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?$\\$ (c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?

Solution :

(a) No. Gravitational influence of matter on nearby objects cannot be screened by any means. This is because gravitational force unlike electrical forces is independent of the nature of the material medium. Also, it is independent of the status of other objects.$\\$ (b) Yes. If the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g).$\\$ (c) Tidal effect depends inversely upon the cube of the distance while, gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull.

2   Choose the correct alternative:$\\$ (a) Acceleration due to gravity increases/decreases with increasing altitude.$\\$ (b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).$\\$ (c) Acceleration due to gravity is independent of mass of the earth/mass of the body.$\\$ (d) The formula $-G Mm(1/r_2-1/r_1)$is more/less accurate than the formula $mg(r_2-r_1)$ for the difference of potential energy between two points $r_2$ and $r_1$ distance away from the centre of the earth.

Solution :

(a) Decreases.$\\$ Explanation : Acceleration due to gravity at depth h is given by the relation:$\\$ $g_h=(1-\dfrac{2h}{r_e})g$$\\$ Where, $\\$ $R_e=$Radius of the Earth\\ $g =$ Acceleration due to gravity on the surface of the Earth$\\$ It is clear from the given relation that acceleration due to gravity decreases with an increase in height.$\\$ (b) Decreases.$\\$ Explanation: Acceleration due to gravity at depth d is given by the relation:$\\$ $g_d=(1-\dfrac{d}{R_e})g$$\\$ It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.$\\$ (c) Mass of the body.$\\$ Explanation: Acceleration due to gravity of body of mass $m$ is given by the relation:$\\$ $g=\dfrac{GM}{R^2}$$\\$ Where,$\\$ $G =$ Universal gravitational constant$\\$ $M =$ Mass of the Earth$\\$ $R =$ Radius of the Earth$\\$ Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.$\\$ (d) More.$\\$ Explanation: Gravitational potential energy of two points $r_2 $ and $r_1$ distance away from the centre of the Earth is respectively given by:$\\$ $V_{r1}=-\dfrac{GmM}{r_1}\\ V_{r2}=-\dfrac{GmM}{r_2}\\ \\ \\ \\ V(r_1)=\dfrac{GmM}{r_1}\\ V(r_2)=\dfrac{GmM}{r_2}$$\\$ $\therefore \text{ Differencein potential energy,} V=V(r_2)-V(r_1)\\ =-GmM(\dfrac{1}{r_2}-\dfrac{1}{r_1})$$\\$ Hence, this formula is more accurate than the formula $mg ( r_2 - r_1 ).$

3   Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

Solution :

Time taken by the Earth to complete one revolution around the Sun,$\\$ $T_e=1\text{year}$$\\$ Orbital radius of the Earth in its orbit, $R_e = 1 AU$$\\$ Time taken by the planet to complete one revolution around the Sun,$\\$ $T_p=\dfrac{1}{2}T_e=\dfrac{1}{2}\text{year}$$\\$ Orbital radius of the planet = $R_p$$\\$ From Kepler’s third law of planetary motion, we can write:$\\$ $(\dfrac{R_p}{R_e})^3=(\dfrac{T_p}{T_e})^2\\ \dfrac{R_p}{R_e}=(\dfrac{T_p}{T_e})^{\frac{2}{3}}\\ =(\dfrac{\dfrac{1}{2}}{1})^{\frac{2}{3}}\\ =(o.5)^{\frac{2}{3}} =0.63$$\\$ Hence, the orbital radius of the planet will be $0.63$ times smaller than that of the Earth.

4   Io, one of the satellites of Jupiter, has an orbital period of $1.769$ days and the radius of the orbit is $4.22 * 10 8 m$ . Show that the mass of Jupiter is about one-thousandth that of the sun.

Solution :

Orbital period of Io, $I_0,T_{10}=1.769 \text{days}$$\\$ $=1.769*24*60*60 s$$\\$ Orbital radius of Io,$I_0,R_{10}=4.22*10^8 m$$\\$ Satellite $I _0$ is revolving around the Jupiter$\\$ Mass of the Jupiter is given by the relation:$\\$ $M_j=\dfrac{4\pi^2R^3_{10}}{GT^2_{10}} ..(i)$$\\$ Where,$\\$ $M_j=\text{Mass of Jupiter}\\ G=\text{Universal gravitational constant }$$\\$ Orbital period of Earth,$\\$ $R_e=1AU=1.496*10^{11}m$$\\$ Mass of sun is given as:$\\$ $M_s=\dfrac{4\pi^2R^3_e}{GT^2_e}....(ii)\\ \therefore \dfrac{M_x}{M_j}=\dfrac{4 \pi^2 R^3_e}{GT^2_e}*\dfrac{GT^2_{10}}{4\pi^2R^3_{10}}\\ =\dfrac{R^3_e}{R^3_{10}}*\dfrac{T^2_{10}}{T^2_e}\\ =(\dfrac{1.769*24*60*60}{365.25*24*60*60})^2\\ *(\dfrac{1.496*10^{11}}{4.22*10^8})^3\\ =1045.04\\ \therefore \dfrac{M_s}{M_j}\sim 1000\\ M_s \sim 1000*M_j\\ \therefore \dfrac{M_s}{M_j}=\dfrac{4\pi^2R^3_e}{GT^2_e}*\dfrac{GT^2_{10}}{4 \pi^2R^3_{10}}\\ \dfrac{M_s}{M_j}= \dfrac{R^3_e}{T^2_e}*\dfrac{T^2_{10}}{R^3_{10}}\\ =\dfrac{(1.496*10^{11})^3}{(365.25*24*60*60)^2}\\ *\dfrac{(1.769*24*60*60)^2}{(4.22*10^8)^3}\\ \dfrac{M_s}{M_j}=1045.04\approx 1000\\ M_s=1000*M_j$$\\$ Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.

5   Let us assume that our galaxy consists of $2.5 * 10 ^{11}$ stars each of one solar mass. How long will a star at a distance of $50,000$ ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be $10^ 5$ ly .

Solution :

Mass of our galaxy Milky Way, $M=2.5*10^{11}$ solar mass$\\$ Solar mass =Mass of Sun=$2.0*10^{36} kg $$\\$ Mass of our galaxy,$M=2.5 *10^{11}*2*10^{36} \\ =5*10^{41} kg $$\\$ Diameter of Milky Way, $d = 10 ^5 ly$$\\$ Radius of Milky Way, $r =5 * 10^ 4 l y$$\\$ $1 ly = 9.46 * 10^{15}m\\ \therefore r=5*10^{4}* 9.46* 10^{15}\\ =4.73* 10^{20}m$$\\$ Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:$\\$ $T=(\dfrac{4\pi^2 r^3}{GM})^{\frac{1}{2}}\\ =(\dfrac{4*(3.14)^2*(4.73)^3*10^{60}}{6.67*10^{-11}*5*10^{41}})^{\frac{1}{2}}\\ =(\dfrac{39.48*105.85*10^{30}}{33.35})^{\frac{1}{2}}\\ =(25.27 *10 ^{30})^{\frac{1}{2}} \\ =1.12 * 10^{16} s\\ 1\text{year}=365*324*60*60 s\\ 1s=\dfrac{1}{365 * 24 * 60 * 60}\text{year}\\ \therefore 1.12 * 10^{16} s\\ =\dfrac{1.12*10^{16}}{365 *24 * 60 * 60}\\ 3.55*10^8\text{years}$